%I #13 Jul 23 2017 02:41:17
%S 5,15,21,33,54,54,85,90,100,117,110,135,153,161,204,195,252,261,315,
%T 308,315,315,351,385,405,418,414,450,455,476,510,533,595,609,637,650,
%U 705,693,684,777,792,870,884,931,986,999,1071,1105,1121,1110,1125,1210,1230
%N Numbers S with two runsums (sum of positive integer runs) where the 2 runs are separated by a single number gap (b), i.e., S = a + (a+1) + ... + (b-1) = (b+1) + ... + c with a < b-1, b < c.
%C The sums are the difference of two triangular numbers A000217. The sum series where the missing number is included is A110701. The numbers in the gap between the two runs are A094550.
%H Ron Knott <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/">Runsums</a>
%H T. Verhoeff, <a href="https://cs.uwaterloo.ca/journals/JIS/trapzoid.html">Rectangular and Trapezoidal Arrangements</a>, J. Integer Sequences, Vol. 2, 1999, #99.1.6.
%e 1+2+3+4+5 and 7+8 have a gap between them of (6). The sum of both is 15 so 15 is in the ordered series (as a(2)).
%Y Cf. A001227, A094550, A110701, A110703.
%K nonn
%O 1,1
%A _Ron Knott_, Aug 04 2005